Question: A particle moves along the curve $xy=16$ so that that the $y$ -coordinate is increasing at a constant rate of $2$ units per minute. What is the magnitude (in units per minute) of the particle's velocity vector when the particle is at the point $(4,4)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $4\sqrt{2}$ (Choice C) C $8$ (Choice D) D $2\sqrt{2}$
Explanation: Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This however shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dy}{dt}=2$ for any value of $t$. We are asked for the magnitude of the particle's velocity vector when the particle is at the point $(4,4)$. In other words, we are asked for $\left|\left|\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)\right|\right|$ at the point $(4,4)$. Finding $\dfrac{dx}{dt}$ $\dfrac{dx}{dt}=-\dfrac{2x}{y}$ Finding $\dfrac{dx}{dt}$ at $(4,4)$ The expression for $\dfrac{dx}{dt}$ depends on both the particle's $x$ -coordinate ${4}$ and its $y$ -coordinate ${4}$ : $\begin{aligned} \dfrac{dx}{dt}&=-\dfrac{2({4})}{({4})} \\\\ &=-2 \end{aligned}$ Therefore, the particle's velocity vector at the point $(4,4)$ is $(-2,2)$. Finding $||(-2,2)||$ $||(-2,2)||=2\sqrt{2}$ In conclusion, the magnitude of the particle's velocity vector when the particle is at the point $(4,4)$ is $2\sqrt{2}$ units per minute.